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Shtirlitz [24]
2 years ago
6

Please help!

Mathematics
1 answer:
jasenka [17]2 years ago
8 0
0%. You can't pick a 3 then pick a 3 again, as there is only one 3 available to be picked. 0/1
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What is the equation of the line that is perpendicular to the given line and passes through the point (2, 6)?
natali 33 [55]

Answer:

Either x = 2 or y = 6, depending on the original line

Step-by-step explanation:

So, if the original line is horizontal, our new line is vertical, and all vertical lines in a graph is x = some number. To pass through the point (2, 6), x has to equal 2, since the point's x-coordinate is 2.

On the other hand, if the original line is vertical, our new line is horizontal, which is y = some number. Our point's y-coordinate is 6, so our line should be that y = 6.

5 0
3 years ago
What s the equation for y= (0,2) slope= -3/7
masya89 [10]
The standard form equation for a straight line is y=mx+b where m is the slope and b is the y intercept. Substitute your given information:
y= -3/7x+2 is the equation
5 0
3 years ago
___,36 ,108 ,324 ,972
hjlf
12

First figure out of it is a arithmetic or geometric sequence, in this case it is a geometric sequence because you have to multiply, not add to find the next number.

Divide one number, I'll pick 324 by the number before it, 108, you get three.

Now divide 36 by 3, you get 12.

To check, multiply 12 by 3, you get 36, multiply that by 3, you get 108, and so on and so forth.
3 0
3 years ago
What is the opposite of -63
Evgesh-ka [11]

Answer:

63

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the exact value of cos(sin^-1(-5/13))
son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0
2 years ago
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